Question

The number of solutions of the equation $(\log_2 \cos \theta)^2 + \log_{\cos \theta}^4 (16 \cos \theta) = 2$, in interval $[0, 2\pi)$ is

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (C)

2

Answer 2

Let the given equation be $(\log_2 \cos \theta)^2 + \log_{\cos \theta}^4 (16 \cos \theta) = 2$. The domain requires $\cos \theta > 0$ and $\cos \theta \neq 1$. In $[0, 2\pi)$, this means $\theta \in (0, \pi/2) \cup (3\pi/2, 2\pi)$, which implies $0 < \cos \theta < 1$.

Assuming the notation $\log_{\cos \theta}^4 (16 \cos \theta)$ means $\log_{(\cos \theta)^4} (16 \cos \theta)$: Let $y = \log_2 \cos \theta$. Since $0 < \cos \theta < 1$, we have $y < 0$. The equation becomes $y^2 + \log_{(\cos \theta)^4} (16 \cos \theta) = 2$. Using the change of base formula: $\log_{(\cos \theta)^4} (16 \cos \theta) = \frac{\log_2 (16 \cos \theta)}{\log_2 ((\cos \theta)^4)} = \frac{\log_2 16 + \log_2 \cos \theta}{4 \log_2 \cos \theta} = \frac{4+y}{4y}$. The equation is $y^2 + \frac{4+y}{4y} = 2$. $y^2 + \frac{1}{y} + \frac{1}{4} = 2$ $y^2 + \frac{1}{y} - \frac{7}{4} = 0$ Multiplying by $4y$: $4y^3 + 4 - 7y = 0 \implies 4y^3 - 7y + 4 = 0$. Let $f(y) = 4y^3 - 7y + 4$. We need roots where $y < 0$. $f'(y) = 12y^2 - 7$. For $y < 0$, $f'(y) = 0$ at $y = -\sqrt{7/12}$. $f(0) = 4$. $f(-2) = 4(-8) - 7(-2) + 4 = -32 + 14 + 4 = -14$. Since $f(-2) < 0$ and $f(0) > 0$, there is a root $y_0 \in (-2, 0)$. The local maximum at $y = -\sqrt{7/12}$ is $f(-\sqrt{7/12}) = (14/3)\sqrt{7/12} + 4 > 0$. As $y \to -\infty$, $f(y) \to -\infty$. Thus, there is exactly one negative root $y_0$. So, $\log_2 \cos \theta = y_0$, which means $\cos \theta = 2^{y_0}$. Since $y_0 \in (-2, 0)$, $2^{y_0} \in (2^{-2}, 2^0) = (1/4, 1)$. For any $c \in (0, 1)$, $\cos \theta = c$ has two solutions in $[0, 2\pi)$. Thus, there are 2 solutions.

Assuming the notation $\log_{\cos \theta}^4 (16 \cos \theta)$ means $\log_{\cos \theta} ((16 \cos \theta)^4)$: The equation becomes $y^2 + \log_{\cos \theta} ((16 \cos \theta)^4) = 2$. $y^2 + \frac{4 \log_2 (16 \cos \theta)}{\log_2 \cos \theta} = 2$ $y^2 + \frac{4(4+y)}{y} = 2$ $y^2 + \frac{16}{y} + 4 = 2$ $y^2 + \frac{16}{y} + 2 = 0$ Multiplying by $y$: $y^3 + 16 + 2y = 0 \implies y^3 + 2y + 16 = 0$. Let $g(y) = y^3 + 2y + 16$. $g'(y) = 3y^2 + 2 > 0$, so $g(y)$ is strictly increasing. $g(-2) = -8 - 4 + 16 = 4$. $g(-3) = -27 - 6 + 16 = -17$. There is one real root $y_0 \in (-3, -2)$. So, $\log_2 \cos \theta = y_0$, which means $\cos \theta = 2^{y_0}$. Since $y_0 \in (-3, -2)$, $2^{y_0} \in (2^{-3}, 2^{-2}) = (1/8, 1/4)$. For any $c \in (0, 1)$, $\cos \theta = c$ has two solutions in $[0, 2\pi)$. Thus, there are 2 solutions.

Both plausible interpretations lead to 2 solutions.